Overview

This assignment provides an introdcution to using the prioritizr R package for systematic conservation planning.

if (!require("librarian")){
  install.packages("librarian")
  library(librarian)
}

## Loading required package: librarian

librarian::shelf(
  assertthat, BiocManager, dplyr, gridExtra, here, mapview, 
  prioritizr, prioritizrdata, 
  raster, remotes, rgeos, rgdal, scales, sf, sp, stringr,
  units)

## 
##   The 'cran_repo' argument in shelf() was not set, so it will use
##   cran_repo = 'https://cran.r-project.org' by default.
## 
##   To avoid this message, set the 'cran_repo' argument to a CRAN
##   mirror URL (see https://cran.r-project.org/mirrors.html) or set
##   'quiet = TRUE'.

## Warning: multiple methods tables found for 'crop'

## Warning: multiple methods tables found for 'extend'

if (!require("lpsymphony")){
  BiocManager::install("lpsymphony")
  library(lpsymphony)
}

## Loading required package: lpsymphony

dir_data <- here("data/prioritizr")
pu_shp   <- file.path(dir_data, "pu.shp")
pu_url   <- "https://github.com/prioritizr/massey-workshop/raw/main/data.zip"
pu_zip   <- file.path(dir_data, basename(pu_url))
vegetation_tif <- file.path(dir_data, "vegetation.tif")

dir.create(dir_data, showWarnings = F, recursive = T)
if (!file.exists(pu_shp)){
  download.file(pu_url, pu_zip)
  unzip(pu_zip, exdir = dir_data)
  dir_unzip   <- file.path(dir_data, "data")
  files_unzip <- list.files(dir_unzip, full.names = T)
  file.rename(
    files_unzip, 
    files_unzip %>% str_replace("prioritizr/data", "prioritizr"))
  unlink(c(pu_zip, dir_unzip), recursive = T)
}

1 Data

This data was obtained from the “Introduction to Marxan” course and was originally a subset of a larger spatial prioritization project performed under contract to Australia’s Department of Environment and Water Resources. It contains vector-based planning unit data and the raster-based data describing the spatial distributions of vegetation classes in southern Tasmania, Australia.

2.1 Data import

# import planning unit data
pu_data <- as(read_sf(pu_shp), "Spatial")

# format columns in planning unit data
pu_data$locked_in <- as.logical(pu_data$locked_in)
pu_data$locked_out <- as.logical(pu_data$locked_out)

# import vegetation data
veg_data <- stack(vegetation_tif)

2.2 Planning unit data

The planning unit data contains spatial data describing the geometry for each planning unit and attribute data with information about each planning unit (e.g. cost values). The attribute data contains 5 columns with contain the following information:

id: unique identifiers for each planning unit
cost: acquisition cost values for each planning unit (millions of Australian dollars).
status: status information for each planning unit (only relevant with Marxan)
locked_in: logical values (i.e. TRUE/FALSE) indicating if planning units are covered by protected areas or not.
locked_out: logical values (i.e. TRUE/FALSE) indicating if planning units cannot be managed as a protected area because they contain are too degraded.

# print a short summary of the data
print(pu_data)

## class       : SpatialPolygonsDataFrame 
## features    : 516 
## extent      : 348703.2, 611932.4, 5167775, 5344516  (xmin, xmax, ymin, ymax)
## crs         : +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs 
## variables   : 5
## names       :   id,             cost, status, locked_in, locked_out 
## min values  :  557, 3.59717531470679,      0,         0,          0 
## max values  : 1130,  47.238336402701,      2,         1,          1

# plot the planning unit data
plot(pu_data)

# plot an interactive map of the planning unit data
mapview(pu_data)

Explore the planning unit data

# print the structure of object
str(pu_data, max.level = 2)

## Formal class 'SpatialPolygonsDataFrame' [package "sp"] with 5 slots
##   ..@ data       :'data.frame':  516 obs. of  5 variables:
##   ..@ polygons   :List of 516
##   ..@ plotOrder  : int [1:516] 69 104 1 122 157 190 4 221 17 140 ...
##   ..@ bbox       : num [1:2, 1:2] 348703 5167775 611932 5344516
##   .. ..- attr(*, "dimnames")=List of 2
##   ..@ proj4string:Formal class 'CRS' [package "sp"] with 1 slot

# print the class of the object
class(pu_data)

## [1] "SpatialPolygonsDataFrame"
## attr(,"package")
## [1] "sp"

# print the slots of the object
slotNames(pu_data)

## [1] "data"        "polygons"    "plotOrder"   "bbox"        "proj4string"

# print the coordinate reference system
print(pu_data@proj4string)

## CRS arguments:
##  +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs

# print number of planning units (geometries) in the data
nrow(pu_data)

## [1] 516

# print the first six rows in the data
head(pu_data@data)

##    id     cost status locked_in locked_out
## 1 557 29.74225      0     FALSE      FALSE
## 2 558 29.87703      0     FALSE      FALSE
## 3 574 28.60687      0     FALSE      FALSE
## 4 575 30.83416      0     FALSE      FALSE
## 5 576 38.75511      0     FALSE      FALSE
## 6 577 38.11618      2      TRUE      FALSE

# print the first six values in the cost column of the attribute data
head(pu_data$cost)

## [1] 29.74225 29.87703 28.60687 30.83416 38.75511 38.11618

# print the highest cost value
max(pu_data$cost)

## [1] 47.23834

# print the smallest cost value
min(pu_data$cost)

## [1] 3.597175

# print average cost value
mean(pu_data$cost)

## [1] 26.87393

# plot a map of the planning unit cost data
spplot(pu_data, "cost")

# plot an interactive map of the planning unit cost data
mapview(pu_data, zcol = "cost")

2.2 Questions

How many planning units are in the planning unit data?

Answer

What is the highest cost value?

Answer

Is there a spatial pattern in the planning unit cost values (hint: use plot to make a map)?

Answer

2.3 Vegetation data

The vegetation data describe the spatial distribution of 32 vegetation classes in the study area. This data is in a raster format and so the data are organized using a grid comprising square grid cells that are each the same size.

# print a short summary of the data
print(veg_data)

## class      : RasterStack 
## dimensions : 164, 326, 53464, 32  (nrow, ncol, ncell, nlayers)
## resolution : 967, 1020  (x, y)
## extent     : 298636.7, 613878.7, 5167756, 5335036  (xmin, xmax, ymin, ymax)
## crs        : +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs 
## names      : vegetation.1, vegetation.2, vegetation.3, vegetation.4, vegetation.5, vegetation.6, vegetation.7, vegetation.8, vegetation.9, vegetation.10, vegetation.11, vegetation.12, vegetation.13, vegetation.14, vegetation.15, ... 
## min values :            0,            0,            0,            0,            0,            0,            0,            0,            0,             0,             0,             0,             0,             0,             0, ... 
## max values :            1,            1,            1,            1,            1,            1,            1,            1,            1,             1,             1,             1,             1,             1,             1, ...

# plot a map of the 20th vegetation class
plot(veg_data[[20]])

# plot an interactive map of the 20th vegetation class
mapview(veg_data[[20]])

Preview feature data

# print number of rows in the data
nrow(veg_data)

## [1] 164

# print number of columns  in the data
ncol(veg_data)

## [1] 326

# print number of cells in the data
ncell(veg_data)

## [1] 53464

# print number of layers in the data
nlayers(veg_data)

## [1] 32

# print resolution on the x-axis
xres(veg_data)

## [1] 967

# print resolution on the y-axis
yres(veg_data)

## [1] 1020

# print spatial extent of the grid, i.e. coordinates for corners
extent(veg_data)

## class      : Extent 
## xmin       : 298636.7 
## xmax       : 613878.7 
## ymin       : 5167756 
## ymax       : 5335036

# print the coordinate reference system
print(veg_data@crs)

## CRS arguments:
##  +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs

# print a summary of the first layer in the stack
print(veg_data[[1]])

## class      : RasterLayer 
## band       : 1  (of  32  bands)
## dimensions : 164, 326, 53464  (nrow, ncol, ncell)
## resolution : 967, 1020  (x, y)
## extent     : 298636.7, 613878.7, 5167756, 5335036  (xmin, xmax, ymin, ymax)
## crs        : +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs 
## source     : vegetation.tif 
## names      : vegetation.1 
## values     : 0, 1  (min, max)

# print the value in the 800th cell in the first layer of the stack
print(veg_data[[1]][800])

##   
## 0

# print the value of the cell located in the 30th row and the 60th column of
# the first layer
print(veg_data[[1]][30, 60])

##   
## 0

# calculate the sum of all the cell values in the first layer
cellStats(veg_data[[1]], "sum")

## [1] 17

# calculate the maximum value of all the cell values in the first layer
cellStats(veg_data[[1]], "max")

## [1] 1

# calculate the minimum value of all the cell values in the first layer
cellStats(veg_data[[1]], "min")

## [1] 0

# calculate the mean value of all the cell values in the first layer
cellStats(veg_data[[1]], "mean")

## [1] 0.00035239

2.3 Questions

What part of the study area is the 13th vegetation class found in (hint: make a map)? For instance, is it in the south-eastern part of the study area?

Answer

What proportion of cells contain the 12th vegetation class?

Answer

Which vegetation class is the most abundant (i.e. present in the greatest number of cells)?

Answer

3 Gap analysis

3.1 Introduction

A gap analysis involves calculating how well each of our biodiversity features (i.e. vegetation classes in this exercise) are represented (covered) by protected areas. Next, the current representation by protected areas of each feature (e.g. 5% of their spatial distribution covered by protected areas) is compared to a target threshold (e.g. 20% of their spatial distribution covered by protected areas). This target threshold denotes the minimum amount (e.g. minimum proportion of spatial distribution) that we need of each feature to be represented in the protected area system. Ideally, targets should be based on an estimate of how much area or habitat is needed for ecosystem function or species persistence. In practice, targets are generally set using simple rules of thumb (e.g. 10% or 20%), policy, or standard practices (e.g. setting targets for species based on geographic range size).

3.2 Feature abundance

This section is to calculate how much of each vegetation feature occurs inside each planning unit (i.e. the abundance of the features). The problem function creates an empty conservation planning problem that only contains the planning unit and biodiversity data. Then the feature_abundances function is used to calculate the total amount of each feature in each planning unit.

# create prioritizr problem with only the data
p0 <- problem(pu_data, veg_data, cost_column = "cost")

# print empty problem,
# we can see that only the cost and feature data are defined
print(p0)

## Conservation Problem
##   planning units: SpatialPolygonsDataFrame (516 units)
##   cost:           min: 3.59718, max: 47.23834
##   features:       vegetation.1, vegetation.2, vegetation.3, ... (32 features)
##   objective:      none
##   targets:        none
##   decisions:      default
##   constraints:    <none>
##   penalties:      <none>
##   portfolio:      default
##   solver:         default

# calculate amount of each feature in each planning unit
abundance_data <- feature_abundances(p0)

# print abundance data
print(abundance_data)

## # A tibble: 32 × 3
##    feature       absolute_abundance relative_abundance
##    <chr>                      <dbl>              <dbl>
##  1 vegetation.1                16.0                  1
##  2 vegetation.2                14.3                  1
##  3 vegetation.3                10.4                  1
##  4 vegetation.4                17.8                  1
##  5 vegetation.5                13.0                  1
##  6 vegetation.6                14.3                  1
##  7 vegetation.7                20.0                  1
##  8 vegetation.8                14.0                  1
##  9 vegetation.9                18.0                  1
## 10 vegetation.10               20.0                  1
## # … with 22 more rows

# note that only the first ten rows are printed,
# this is because the abundance_data object is a tibble (i.e. tbl_df) object
# and not a standard data.frame object
print(class(abundance_data))

## [1] "tbl_df"     "tbl"        "data.frame"

# print all of the rows in abundance_data like this
print(abundance_data, n = Inf)

## # A tibble: 32 × 3
##    feature       absolute_abundance relative_abundance
##    <chr>                      <dbl>              <dbl>
##  1 vegetation.1                16.0                  1
##  2 vegetation.2                14.3                  1
##  3 vegetation.3                10.4                  1
##  4 vegetation.4                17.8                  1
##  5 vegetation.5                13.0                  1
##  6 vegetation.6                14.3                  1
##  7 vegetation.7                20.0                  1
##  8 vegetation.8                14.0                  1
##  9 vegetation.9                18.0                  1
## 10 vegetation.10               20.0                  1
## 11 vegetation.11               23.6                  1
## 12 vegetation.12              748.                   1
## 13 vegetation.13              126.                   1
## 14 vegetation.14               10.5                  1
## 15 vegetation.15               17.5                  1
## 16 vegetation.16               15.0                  1
## 17 vegetation.17              213.                   1
## 18 vegetation.18               14.3                  1
## 19 vegetation.19               17.1                  1
## 20 vegetation.20               21.4                  1
## 21 vegetation.21               18.6                  1
## 22 vegetation.22              297.                   1
## 23 vegetation.23               20.3                  1
## 24 vegetation.24              165.                   1
## 25 vegetation.25              716.                   1
## 26 vegetation.26               24.0                  1
## 27 vegetation.27               18.8                  1
## 28 vegetation.28               17.5                  1
## 29 vegetation.29               24.7                  1
## 30 vegetation.30               59.0                  1
## 31 vegetation.31               60.0                  1
## 32 vegetation.32               32                    1

The abundance_data object contains three columns:
- The feature column contains the name of each feature (derived from names(veg_data)).
- The absolute_abundance column contains the total amount of each feature in all the planning units.
- The relative_abundance column contains the total amount of each feature in the planning units expressed as a proportion of the total amount in the underlying raster data.
Since all the raster cells containing vegetation overlap with the planning units, all of the values in the relative_abundance column are equal to one (meaning 100%). So the relative_abundance per feature is a measure of the ‘percent presence’ of that feature across all planning units (100% or 1 in the case of all these vegetation layers, which is not interesting), whereas absolute_abundance measures the total amount of that feature when the value for all planning units is added up.
Now add a new column with the feature abundances expressed in area units (i.e. km²).

# add new column with feature abundances in km^2
abundance_data$absolute_abundance_km2 <-
  (abundance_data$absolute_abundance * prod(res(veg_data))) %>%
  set_units(m^2) %>%
  set_units(km^2)

# print abundance data
print(abundance_data)

## # A tibble: 32 × 4
##    feature       absolute_abundance relative_abundance absolute_abundance_km2
##    <chr>                      <dbl>              <dbl>                 [km^2]
##  1 vegetation.1                16.0                  1                   15.8
##  2 vegetation.2                14.3                  1                   14.1
##  3 vegetation.3                10.4                  1                   10.2
##  4 vegetation.4                17.8                  1                   17.6
##  5 vegetation.5                13.0                  1                   12.8
##  6 vegetation.6                14.3                  1                   14.1
##  7 vegetation.7                20.0                  1                   19.7
##  8 vegetation.8                14.0                  1                   13.9
##  9 vegetation.9                18.0                  1                   17.8
## 10 vegetation.10               20.0                  1                   19.7
## # … with 22 more rows

# calculate the average abundance of the features
mean(abundance_data$absolute_abundance_km2)

## 86.82948 [km^2]

# plot histogram of the features' abundances
hist(abundance_data$absolute_abundance_km2, main = "Feature abundances")

# find the abundance of the feature with the largest abundance
max(abundance_data$absolute_abundance_km2)

## 737.982 [km^2]

# find the name of the feature with the largest abundance
abundance_data$feature[which.max(abundance_data$absolute_abundance_km2)]

## [1] "vegetation.12"

3.2 Questions

What is the median abundance of the features (hint: median)?

Answer

What is the name of the feature with smallest abundance?

Answer

How many features have a total abundance greater than 100 km^2 (hint: use sum(abundance_data$absolute_abundance_km2 > set_units(threshold, km^2) with the correct threshold value)?

Answer

3.3 Feature representation

Now calculate the amount of each feature in the planning units that are covered by protected areas (i.e. feature representation by protected areas). This can be done using the eval_feature_representation_summary() function. This function requires: | (i) a conservation problem object with the planning unit and biodiversity data | (ii) an object representing a solution to the problem (i.e an object in the same format as the planning unit data with values indicating if the planning units are selected or not).

# create column in planning unit data with binary values (zeros and ones)
# indicating if a planning unit is covered by protected areas or not
pu_data$pa_status <- as.numeric(pu_data$locked_in)

# calculate feature representation by protected areas
repr_data <- eval_feature_representation_summary(p0, pu_data[, "pa_status"])

# print feature representation data
print(repr_data)

## # A tibble: 32 × 5
##    summary feature       total_amount absolute_held relative_held
##    <chr>   <chr>                <dbl>         <dbl>         <dbl>
##  1 overall vegetation.1          16.0         0            0     
##  2 overall vegetation.2          14.3         0            0     
##  3 overall vegetation.3          10.4         0            0     
##  4 overall vegetation.4          17.8         0            0     
##  5 overall vegetation.5          13.0         0            0     
##  6 overall vegetation.6          14.3         0            0     
##  7 overall vegetation.7          20.0         0            0     
##  8 overall vegetation.8          14.0         0            0     
##  9 overall vegetation.9          18.0         0.846        0.0470
## 10 overall vegetation.10         20.0         0            0     
## # … with 22 more rows

The repr_data object contains three columns:
- The feature column contains the name of each feature.
- The absolute_held column shows the total amount of each feature held in the solution (i.e. the planning units covered by protected areas).
- The relative_held column shows the proportion of each feature held in the solution (i.e. the proportion of each feature’s spatial distribution held in protected areas).
So the absolute_held is an amount up to but not exceeding the original absolute_abundance of that feature across all planning units (see above) based on those planning units in the solution, and the relative_held is like the percent of planning units in the solution that had this feature present. Since the absolute_held values correspond to the number of grid cells in the veg_data object with overlap with protected areas, let’s convert them to area units (i.e. km²) so we can report them.

# add new column with the areas represented in km^2
repr_data$absolute_held_km2 <-
  (repr_data$absolute_held * prod(res(veg_data))) %>%
  set_units(m^2) %>%
  set_units(km^2)

# print representation data
print(repr_data)

## # A tibble: 32 × 6
##    summary feature       total_amount absolute_held relative_held absolute_held_k…
##    <chr>   <chr>                <dbl>         <dbl>         <dbl>           [km^2]
##  1 overall vegetation.1          16.0         0            0                 0    
##  2 overall vegetation.2          14.3         0            0                 0    
##  3 overall vegetation.3          10.4         0            0                 0    
##  4 overall vegetation.4          17.8         0            0                 0    
##  5 overall vegetation.5          13.0         0            0                 0    
##  6 overall vegetation.6          14.3         0            0                 0    
##  7 overall vegetation.7          20.0         0            0                 0    
##  8 overall vegetation.8          14.0         0            0                 0    
##  9 overall vegetation.9          18.0         0.846        0.0470            0.834
## 10 overall vegetation.10         20.0         0            0                 0    
## # … with 22 more rows

3.3 Questions

What is the average proportion of the features held in protected areas (hint: use mean(table$relative_held) with the correct table name)?

Answers

If we set a target of 10% coverage by protected areas, how many features fail to meet this target (hint: use sum(table$relative_held >= target_value) with the correct table name)?

Answers

If we set a target of 20% coverage by protected areas, how many features fail to meet this target?

Answers

Is there a relationship between the total abundance of a feature and how well it is represented by protected areas (hint: plot(abundance_data$absolute_abundance ~ repr_data$relative_held))?

Answers

4. Spatial prioritizations

4.1. Introduction

Prioritizations are developed to identify priority areas for protected area establishment. Its worth noting that prioritizr is a decision support tool (similar to Marxan and Zonation).

4.2 Simple Prioritization

Create a prioritization using the minimum set formulation of the reserve selection problem. This formulation means that will meet the targets for biodiversity features for minimum cost. Here, we will set 5% targets for each vegetation class and use the data in the cost column to specify acquisition costs.

# print planning unit data
print(pu_data)

## class       : SpatialPolygonsDataFrame 
## features    : 516 
## extent      : 348703.2, 611932.4, 5167775, 5344516  (xmin, xmax, ymin, ymax)
## crs         : +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs 
## variables   : 6
## names       :   id,             cost, status, locked_in, locked_out, pa_status 
## min values  :  557, 3.59717531470679,      0,         0,          0,         0 
## max values  : 1130,  47.238336402701,      2,         1,          1,         1

# make prioritization problem
p1_rds <- file.path(dir_data, "p1.rds")
if (!file.exists(p1_rds)){
  p1 <- problem(pu_data, veg_data, cost_column = "cost") %>%
        add_min_set_objective() %>%
        add_relative_targets(0.05) %>% # 5% representation targets
        add_binary_decisions() %>%
        add_lpsymphony_solver()
  saveRDS(p1, p1_rds)
}
p1 <- readRDS(p1_rds)

# print problem
print(p1)

## Conservation Problem
##   planning units: SpatialPolygonsDataFrame (516 units)
##   cost:           min: 3.59718, max: 47.23834
##   features:       vegetation.1, vegetation.2, vegetation.3, ... (32 features)
##   objective:      Minimum set objective 
##   targets:        Relative targets [targets (min: 0.05, max: 0.05)]
##   decisions:      Binary decision 
##   constraints:    <none>
##   penalties:      <none>
##   portfolio:      default
##   solver:         Lpsymphony [first_feasible (0), gap (0.1), time_limit (2147483647), verbose (1)]

# solve problem
s1 <- solve(p1)

# print solution, the solution_1 column contains the solution values
# indicating if a planning unit is (1) selected or (0) not
print(s1)

## class       : SpatialPolygonsDataFrame 
## features    : 516 
## extent      : 348703.2, 611932.4, 5167775, 5344516  (xmin, xmax, ymin, ymax)
## crs         : +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs 
## variables   : 7
## names       :   id,             cost, status, locked_in, locked_out, pa_status, solution_1 
## min values  :  557, 3.59717531470679,      0,         0,          0,         0,          0 
## max values  : 1130,  47.238336402701,      2,         1,          1,         1,          1

# calculate number of planning units selected in the prioritization
eval_n_summary(p1, s1[, "solution_1"])

## # A tibble: 1 × 2
##   summary  cost
##   <chr>   <dbl>
## 1 overall    15

# calculate total cost of the prioritization
eval_cost_summary(p1, s1[, "solution_1"])

## # A tibble: 1 × 2
##   summary  cost
##   <chr>   <dbl>
## 1 overall  385.

# plot solution
# selected = green, not selected = grey
spplot(s1, "solution_1", col.regions = c("grey80", "darkgreen"), main = "s1",
       colorkey = FALSE)

4.2 Questions

How many planing units were selected in the prioritization? What proportion of planning units were selected in the prioritization?

Answer

Is there a pattern in the spatial distribution of the priority areas?

Answer

Can you verify that all of the targets were met in the prioritization (hint: eval_feature_representation_summary(p1, s1[, "solution_1"]))?

Answer

4.3 Adding complexity

Our first prioritization suffers many limitations, so let’s add additional constraints to the problem to make it more useful. First, let’s lock in planing units that are already by covered protected areas. If some vegetation communities are already secured inside existing protected areas, then we might not need to add as many new protected areas to the existing protected area system to meet their targets. Since our planning unit data (pu_da) already contains this information in the locked_in column, we can use this column name to specify which planning units should be locked in.

# plot locked_in data
# TRUE = blue, FALSE = grey
spplot(pu_data, "locked_in", col.regions = c("grey80", "darkblue"),
       main = "locked_in", colorkey = FALSE)

# make prioritization problem
p2_rds <- file.path(dir_data, "p2.rds")
if (!file.exists(p2_rds)){
  p2 <- problem(pu_data, veg_data, cost_column = "cost") %>%
      add_min_set_objective() %>%
      add_relative_targets(0.05) %>%
      add_locked_in_constraints("locked_in") %>%
      add_binary_decisions() %>%
      add_lpsymphony_solver()
  saveRDS(p2, p2_rds)
}

p2 <- readRDS(p2_rds)

# print problem
print(p2)

## Conservation Problem
##   planning units: SpatialPolygonsDataFrame (516 units)
##   cost:           min: 3.59718, max: 47.23834
##   features:       vegetation.1, vegetation.2, vegetation.3, ... (32 features)
##   objective:      Minimum set objective 
##   targets:        Relative targets [targets (min: 0.05, max: 0.05)]
##   decisions:      Binary decision 
##   constraints:    <Locked in planning units [198 locked units]>
##   penalties:      <none>
##   portfolio:      default
##   solver:         Lpsymphony [first_feasible (0), gap (0.1), time_limit (2147483647), verbose (1)]

# solve problem
s2 <- solve(p2)

# plot solution
# selected = green, not selected = grey
spplot(s2, "solution_1", col.regions = c("grey80", "darkgreen"), main = "s2",
       colorkey = FALSE)

Set targets to 10%.

# make prioritization problem
p3_rds <- file.path(dir_data, "p3.rds")
if (!file.exists(p3_rds)){
  p3 <- problem(pu_data, veg_data, cost_column = "cost") %>%
    add_min_set_objective() %>%
    add_relative_targets(0.1) %>%
    add_locked_in_constraints("locked_in") %>%
    add_binary_decisions() %>%
    add_lpsymphony_solver()
  saveRDS(p3, p3_rds)
}
p3 <- readRDS(p3_rds)

# print problem
print(p3)

## Conservation Problem
##   planning units: SpatialPolygonsDataFrame (516 units)
##   cost:           min: 3.59718, max: 47.23834
##   features:       vegetation.1, vegetation.2, vegetation.3, ... (32 features)
##   objective:      Minimum set objective 
##   targets:        Relative targets [targets (min: 0.1, max: 0.1)]
##   decisions:      Binary decision 
##   constraints:    <Locked in planning units [198 locked units]>
##   penalties:      <none>
##   portfolio:      default
##   solver:         Lpsymphony [first_feasible (0), gap (0.1), time_limit (2147483647), verbose (1)]

# solve problem
s3 <- solve(p3)

# plot solution
# selected = green, not selected = grey
spplot(s3, "solution_1", col.regions = c("grey80", "darkgreen"), main = "s3",
       colorkey = FALSE)

Lock out highly degraded areas. Similar to before, this information is present in our planning unit data so we can use the locked_out column name to achieve this.

# plot locked_out data
# TRUE = red, FALSE = grey
spplot(pu_data, "locked_out", col.regions = c("grey80", "darkred"),
       main = "locked_out", colorkey = FALSE)

# make prioritization problem
p4_rds <- file.path(dir_data, "p4.rds")
if (!file.exists(p4_rds)){
  p4 <- problem(pu_data, veg_data, cost_column = "cost") %>%
    add_min_set_objective() %>%
    add_relative_targets(0.1) %>%
    add_locked_in_constraints("locked_in") %>%
    add_locked_out_constraints("locked_out") %>%
    add_binary_decisions() %>%
    add_lpsymphony_solver()
  saveRDS(p4, p4_rds)
}
p4 <- readRDS(p4_rds)

# print problem
print(p4)

## Conservation Problem
##   planning units: SpatialPolygonsDataFrame (516 units)
##   cost:           min: 3.59718, max: 47.23834
##   features:       vegetation.1, vegetation.2, vegetation.3, ... (32 features)
##   objective:      Minimum set objective 
##   targets:        Relative targets [targets (min: 0.1, max: 0.1)]
##   decisions:      Binary decision 
##   constraints:    <Locked in planning units [198 locked units]
##                    Locked out planning units [6 locked units]>
##   penalties:      <none>
##   portfolio:      default
##   solver:         Lpsymphony [first_feasible (0), gap (0.1), time_limit (2147483647), verbose (1)]

# solve problem
s4 <- solve(p4)

# plot solution
# selected = green, not selected = grey
spplot(s4, "solution_1", col.regions = c("grey80", "darkgreen"), main = "s4",
       colorkey = FALSE)

4.3 Questions

What is the cost of the planning units selected in s2, s3, and s4?

Answer

How many planning units are in s2, s3, and s4?

Answer

Do the solutions with more planning units have a greater cost? Why (or why not)?

Answer

Why does the first solution (s1) cost less than the second solution with protected areas locked into the solution (s2)?

Answer

Why does the third solution (s3) cost less than the fourth solution solution with highly degraded areas locked out (s4)?

Answer

4.4 Penalizing fragmentation

Plans for protected area systems should promote connectivity. However, the prioritizations we have made so far have been highly fragmented. To address this issue, we can add penalties to our conservation planning problem to penalize fragmentation. As a rule of thumb, we generally want penalty values between 0.00001 and 0.01. However, finding a useful penalty value requires calibration. The “correct” penalty value depends on the size of the planning units, the main objective values (e.g. cost values), and the effect of fragmentation on biodiversity persistence.

# make prioritization problem
p5_rds <- file.path(dir_data, "p5.rds")
if (!file.exists(p5_rds)){
  p5 <- problem(pu_data, veg_data, cost_column = "cost") %>%
    add_min_set_objective() %>%
    add_boundary_penalties(penalty = 0.001) %>%
    add_relative_targets(0.1) %>%
    add_locked_in_constraints("locked_in") %>%
    add_locked_out_constraints("locked_out") %>%
    add_binary_decisions() %>%
    add_lpsymphony_solver()
  saveRDS(p5, p5_rds)
}

p5 <- readRDS(p5_rds)

# print problem
print(p5)

## Conservation Problem
##   planning units: SpatialPolygonsDataFrame (516 units)
##   cost:           min: 3.59718, max: 47.23834
##   features:       vegetation.1, vegetation.2, vegetation.3, ... (32 features)
##   objective:      Minimum set objective 
##   targets:        Relative targets [targets (min: 0.1, max: 0.1)]
##   decisions:      Binary decision 
##   constraints:    <Locked out planning units [6 locked units]
##                    Locked in planning units [198 locked units]>
##   penalties:      <Boundary penalties [edge factor (min: 0.5, max: 0.5), penalty (0.001), zones]>
##   portfolio:      default
##   solver:         Lpsymphony [first_feasible (0), gap (0.1), time_limit (2147483647), verbose (1)]

# solve problem,
# note this will take a bit longer than the previous runs
s5 <- solve(p5)

# print solution
print(s5)

## class       : SpatialPolygonsDataFrame 
## features    : 516 
## extent      : 348703.2, 611932.4, 5167775, 5344516  (xmin, xmax, ymin, ymax)
## crs         : +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs 
## variables   : 7
## names       :   id,             cost, status, locked_in, locked_out, pa_status, solution_1 
## min values  :  557, 3.59717531470679,      0,         0,          0,         0,          0 
## max values  : 1130,  47.238336402701,      2,         1,          1,         1,          1

# plot solution
# selected = green, not selected = grey
spplot(s5, "solution_1", col.regions = c("grey80", "darkgreen"), main = "s5",
       colorkey = FALSE)

Now let’s compare the solutions to the problems with (s5) and without (s4) the boundary length penalties.

4.4 Questions

What is the cost the fourth (s4) and fifth (s5) solutions? Why does the fifth solution (s5) cost more than the fourth (s4) solution?

Answer

Try setting the penalty value to 0.000000001 (i.e. 1e-9) instead of 0.001. What is the cost of the solution now? Is it different from the fourth solution (s4) (hint: try plotting the solutions to visualize them)? Is this is a useful penalty value? Why (or why not)?

Answer

Try setting the penalty value to 0.5. What is the cost of the solution now? Is it different from the fourth solution (s4) (hint: try plotting the solutions to visualize them)? Is this a useful penalty value? Why (or why not)?

Answer

Lab03 Reserves Planning

Julia Parish

2022/02/09

Overview

1 Data

2.1 Data import

2.2 Planning unit data

Explore the planning unit data

2.2 Questions

2.3 Vegetation data

Preview feature data

2.3 Questions

3 Gap analysis

3.1 Introduction

3.2 Feature abundance

3.2 Questions

3.3 Feature representation

3.3 Questions

4. Spatial prioritizations

4.1. Introduction

4.2 Simple Prioritization

4.2 Questions

4.3 Adding complexity

4.3 Questions

4.4 Penalizing fragmentation

4.4 Questions